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IIT JEE Advanced 2011 Paper 1
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© examsnet.com
Question : 29
Total: 69
Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is
1.78 M
2.00 M
2.05 M
2.22 M
Validate
Solution:
The mass of the solution of water and urea
1000 + 12 = 1012 g
Volume of solution =
Mass
of
solution
Density
of
solution
=
1012
1.15
= 973.9 g/mL
Molarity =
Number
of
moles
Volume
of
solution
in
litres
Number of moles =
120
60
= 2
Molarity =
2
0.974
= 2.05 M
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